Introduction to Mathematical Analysis I - Second Edition

80 3.4 PROPERTIES OF CONTINUOUS FUNCTIONS Figure 3.3: Illustration of the Intermediate Value Theorem. Proof: Define ϕ ( x ) = f ( x ) − γ , x ∈ [ a , b ] . Then ϕ is continuous on [ a , b ] . Moreover, ϕ ( a ) ϕ ( b ) = [ f ( a ) − γ ][ f ( b ) − γ ] < 0 . By Theorem 3.4.7 , there exists c ∈ ( a , b ) such that ϕ ( c ) = 0. This is equivalent to f ( c ) = γ . The proof is now complete. Corollary 3.4.9 Let f : [ a , b ] → R be a continuous function. Let m = min { f ( x ) : x ∈ [ a , b ] } and M = max { f ( x ) : x ∈ [ a , b ] } . Then for every γ ∈ [ m , M ] , there exists c ∈ [ a , b ] such that f ( c ) = γ . Example 3.4.1 We will use the Intermediate Value Theorem to prove that the equation e x = − x has at least one real solution. We will assume known that the exponential function is continuous on R and that e x < 1 for x < 0. First define the function f : R → R by f ( x ) = e x + x . Notice that the given equation has a solution x if and only if f ( x ) = 0. Now, the function f is continuous (as the sum of continuous functions). Moreover, note that f ( − 1 ) = e − 1 +( − 1 ) < 1 − 1 = 0 and f ( 0 ) = 1 > 0. We can now apply the Intermediate Value Theorem to the function f on the interval [ − 1 , 0 ] with γ = 0 to conclude that there is c ∈ [ − 1 , 0 ] such that f ( c ) = 0. The point c is the desired solution to the original equation. Example 3.4.2 We show now that, given n ∈ N , every positive real number has a positive n -th root. Let n ∈ N and let a ∈ R with a > 0. First observe that ( 1 + a ) n ≥ 1 + na > a (see Exercise 1.3.7 ) . Now consider the function f : [ 0 , ∞ ) → R given by f ( x ) = x n . Since f ( 0 ) = 0 and f ( 1 + a ) > a , it follows from the Intermediate Value Theorem that there is x ∈ ( 0 , 1 + a ) such that f ( x ) = a . That is, x n = a , as desired. (We show later in Example 4.3.1 that such an x is unique.) We present below a second proof of Theorem 3.4.8 that does not depend on Theorem 3.4.7 , but, instead, relies on the Nested Intervals Theorem (Theorem 2.3.3 ) . Second Proof of Theorem 3.4.8 : We construct a sequence of nested intervals as follows. Set a 1 = a , b 1 = b , and let I 1 = [ a , b ] . Let c 1 = ( a + b ) / 2. If f ( c 1 ) = γ , we are done. Otherwise, either f ( c 1 ) > γ or f ( c 1 ) < γ .